If the magnitude of the electric field inside a uniformly charged spherical shell is zero then is how potential a non-zero constant equal to the potential of shell itself? How does a non-zero potential exist given that there is no need to do work in moving a charge in forceless field?

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    $\begingroup$ The fact that the field is zero indicates that the potential is constant. So I'm a bit unclear what you are asking. $\endgroup$
    – Jon Custer
    Jul 25, 2018 at 14:24
  • $\begingroup$ I have a small confusion that whether electric field is zero exactly at centre or within shell everywhere. $\endgroup$ Jul 25, 2018 at 16:33
  • $\begingroup$ The electric field is zero throughout the interior of the shell (in other words, there is no force field). The potential at a point in space is a property of that location. The amount of work that has to be done to move a charge $q$ from A to B is equal to $W = q\Delta V$. Since the potential in the interior of the spherical shell does not change (because the field is zero, $E = -\frac{dV}{dx}$), the difference in potential between any two points in the interior is zero; this in other words means that no work is done in moving a charge inside the spherical shell. $\endgroup$
    – Yashas
    Jul 27, 2018 at 12:42
  • $\begingroup$ Your conception of work seems to be wrong. Work done is change in potential energy (in this case; it can also be kinetic energy or other things). The fact that potential is same everywhere inside a charged shell means that you won't have to do work to move charges from one place to another, only because you won't encounter a change in potential energy (V×q). This implies that you don't need to work unless you want to change the potential of something. No change of potential does not mean no potential, particularly for this case. I hope this solves your problem. $\endgroup$
    – BeBlunt
    Aug 10, 2020 at 11:21

2 Answers 2


We can first determine the electric field within the shell using Gauss' law, one of Maxwell's equations. Consider a thin shell of radius $R$ which has total surface charge $Q$. For a spherical Gaussian surface $\Sigma$ within the shell, radius $r$, Gauss' law indicates that

$$ \oint_\Sigma \mathbf{E} \cdot d\mathbf{a} = \frac{Q_{\rm enc}}{\epsilon_0} = 0,$$

since we know that $Q_{\rm enc}$, the charged enclosed by our Gaussian surface, is zero. It follows that if $Q_{\rm enc}$, it must be that $\mathbf{E} = \mathbf{0}.$

Since $\mathbf{E}=\mathbf{0}$, this implies that $V = \rm constant$ because of the relationship $\mathbf{E} = -\nabla V$. That is, the (vector) derivative of a constant is zero. This means that the interior is equipotential everywhere, and it takes no work to move a charge anywhere within the shell.

This is much like how it takes no work (against the gravitational field) to move an object horizontally, since there is no change in $mgh$. Every horizontal position along a certain altitude is at a gravitational equipotential.


The potential is defined relative to the infinity - not relative to the center of the shell.

Since there is no field inside the shell, the potential at any point inside the shell is equal to the potential on the surface of the shell, $V=\frac Q {4\pi\epsilon_0}$.

This is illustrated for a positively charged sphere on the diagram below copied from this Hyperphysics page.

enter image description here

The potential in the infinity is defined as zero and it increases as we move toward a positively charged sphere as a positive work would have to be done moving a positive charge against the electric field produced by the sphere.

Inside the sphere, the field is zero, therefore, no work needs to be done to move the charge inside the sphere and, therefore, the potential there does not change.

  • $\begingroup$ Is this field is microscopic or macroscopic? $\endgroup$ Jul 25, 2018 at 18:50
  • $\begingroup$ If in a microscopic field the Electric field vary from point to point inside shell? $\endgroup$ Jul 25, 2018 at 18:51
  • $\begingroup$ @SRIVISHNUBHARAT I am not sure what you mean by microscopic field. In any case though, there is no field inside the shell. $\endgroup$
    – V.F.
    Jul 25, 2018 at 19:21
  • $\begingroup$ This "field" does not have a real existence, in the sense, you can't "see" it (not yet, as of 2020). It's a theoretical understanding; a framework rather, that serves very helpful in studying how charges physically (not sure if that's the term) effect each other - i.e. understanding the extent of "push" and "pull" effects produced by/on various charges. $\endgroup$
    – BeBlunt
    Aug 10, 2020 at 11:26

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